Simplify; express your answer in exponential form. Assume $k\neq 0, y\neq 0$. $\dfrac{{(k^{-2}y^{2})^{-5}}}{{(k^{5}y^{-3})^{2}}}$
Solution: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(k^{-2}y^{2})^{-5} = (k^{-2})^{-5}(y^{2})^{-5}}$ On the left, we have ${k^{-2}}$ to the exponent ${-5}$ . Now ${-2 \times -5 = 10}$ , so ${(k^{-2})^{-5} = k^{10}}$ Apply the ideas above to simplify the equation. $\dfrac{{(k^{-2}y^{2})^{-5}}}{{(k^{5}y^{-3})^{2}}} = \dfrac{{k^{10}y^{-10}}}{{k^{10}y^{-6}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{10}y^{-10}}}{{k^{10}y^{-6}}} = \dfrac{{k^{10}}}{{k^{10}}} \cdot \dfrac{{y^{-10}}}{{y^{-6}}} = k^{{10} - {10}} \cdot y^{{-10} - {(-6)}} = y^{-4}$